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I find this tricky one. How to calculate the first 50 digits/decimals of the fractional number 1/49? Two of my calculators and MatLab gives different answers so I'm curious, how this is calculated "manual" way such that average person with simple sum and multiplication skills needs no calculator for it.

Method 1:

0,
10/49
0,0
100/49 = **2**,2
0,0**2**
20/49
0,020
200/49 = **4**,4
0,020**4**
40/49
0,02040
400/49 = **8**,8
0,02040**8**
800/49 = **16**,16
0,020408**16**
1600/49 = **32**,32
0,02040816**32**
3200/49 = **64**,64
0,0204081632**64**
6400/49 = **130**,30
0,02040816326**530**...

Multiplications and divisions getting big plus backward processing...

Method 2:

1/49 -> rule of second digit of divisor +1 = 5 multiplier. Numbers ending 9 starts with 1. Writing from right to left:

...1
1*5=5+0
...51
5*5=25+0
...551
5*5=25+2=27
...7551
7*5=35+2=37
...77551
7*5=35+3=38
...877551
8*5=40+3=43
...3877551
3*5=15+4=19
...93877551
9*5=45+1=46
...693877551
6*5=30+4=34
...4693877551
4*5=20+3=23
...34693877551
3*5=15+2=17
...734693877551
7*5=35+1=36
.
.
.
...6122448979591836734693877551
6*5=30+0=30
...306122448979591836734693877551
3*5=15
...5306122448979591836734693877551
5*5=25+1=26
...65306122448979591836734693877551
6*5=30+2=32
...265306122448979591836734693877551
2*5=10+3=13
...3265306122448979591836734693877551
3*5=15+1=16
...63265306122448979591836734693877551
6*5=30+1=31
...163265306122448979591836734693877551
1*5=5+3=8
...8163265306122448979591836734693877551
8*5=40
...408163265306122448979591836734693877551
4*5=20
...20408163265306122448979591836734693877551
2*5=10 ENDS HERE
020408163265306122448979591836734693877551

Pretty good approach while seems to be long written this way. Alternative syntax gives more right to the method:

 0 2 0 4 0 8 1 6 3 2 6 5 3 0 6 1 2 2 4 4 8 9 7 9 5 9 1 8 3 6 7 3 4 6 9 3 8 7 7 5 5 1
1           3 1 1 3 2 1       1 1 2 2 4 4 3 4 2 4   4 1 3 3 1 2 3 4 1 4 3 3 2 2

Method 3:

02
0.02
..04
0.0204
....08
0.020408
......16
0.02040816
........32
0.0204081632
..........64
0.020408163265
...........128
0.02040816326530
.............256
0.020408163265306
..............512
0.02040816326530612
...............1024
0.020408163265306122
................2048
0.0204081632653061224
.................4096
0.02040816326530612244
..................8192
0.0204081632653061224489
...................16384
0.02040816326530612244897

Get complicated because numbers needs to be carried backwards, thus memorization...

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HINT:

$$\frac1{49}=\frac2{100-2}=\frac2{100(1-.02)}=.02\left(1-.02\right)^{-1}$$

$$=.02\left(1+.02+(.02)^2+(.02)^3+(.02)^4+\cdots\right)$$

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A method of some generality is as follows. You can solve $$1/49=a\sum_{n=1}^\infty (10^{-k})^{n}$$ This gives $a=(10^k-1)/49$, so $10^{k}\equiv 1$ mod 49.

Using Euler's totient function or even more precisely the Carmichael function and the fact that 10 is a primitive element in the group of units of integers mod 49, we have $k=\varphi(49)=7\cdot 6=42$.

This gives $$a=020 408 163 265 306 122 448 979 591 836 734 693 877 551$$ and hence $$1/49=0.\overline{020 408 163 265 306 122 448 979 591 836 734 693 877 551}$$

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